A026002 a(n) = T(n,n+2), where T = Delannoy triangle (A008288).
1, 7, 41, 231, 1289, 7183, 40081, 224143, 1256465, 7059735, 39753273, 224298231, 1267854873, 7178461215, 40704778785, 231128079903, 1314016698401, 7478998203943, 42612705597769, 243025194476551, 1387226559025961, 7924982285747247, 45308321651001521
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- Luca Ferrari and Emanuele Munarini, Enumeration of Edges in Some Lattices of Paths, J. Int. Seq. 17 (2014) #14.1.5.
- Ricardo Gómez Aíza, Trees with flowers: A catalog of integer partition and integer composition trees with their asymptotic analysis, arXiv:2402.16111 [math.CO], 2024. See p. 19.
Programs
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GAP
List([1..25], n-> (1/n)*Sum([0..n], k-> k*Binomial(n,k)*Binomial(n+k,k+1) )); # G. C. Greubel, Feb 13 2020
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Magma
[(1/n)*(&+[k*Binomial(n, k)*Binomial(n+k, k+1): k in [0..n]]): n in [1..25]]; // G. C. Greubel, Feb 13 2020
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Maple
a:=n->(1/n)*sum(k*binomial(n,k)*binomial(n+k,k+1),k=0..n): seq(a(n),n=1..25); # Emeric Deutsch
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Mathematica
Table[SeriesCoefficient[1/2-1/(2*x)+(1-4*x+x^2)/(2*x*Sqrt[1-6*x+x^2]),{x,0,n}],{n,1,25}] (* Vaclav Kotesovec, Oct 08 2012 *) a[n_] := Binomial[2*n, n+1]*Hypergeometric2F1[-n+1, -n-1, -2*n, -1]; Flatten[Table[a[n], {n, 1, 22}]] (* Detlef Meya, Dec 26 2023 *)
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PARI
my(x='x+O('x^66)); Vec( 1/2-1/(2*x)+(1-4*x+x^2)/(2*x*sqrt(1-6*x+x^2)) ) \\ Joerg Arndt, May 04 2013
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Sage
a = lambda n: hypergeometric([-n-1, -n+1], [1], 2) [simplify(a(n)) for n in (1..25)] # Peter Luschny, Nov 19 2014
Formula
From Emeric Deutsch, Dec 06 2003: (Start)
a(n) = (1/n)*Sum_{k=0..n} k*binomial(n, k)*binomial(n+k, k+1).
G.f.: 1/2 - 1/(2*z) + (1-4*z+z^2)/(2*z*sqrt(1-6*z+z^2)). (End)
a(n) = Sum_{k=0..floor(n/2)} k*A110220(n, k). - Emeric Deutsch, Jul 16 2005
a(n) = Sum_{k=0..n} C(n, k)*C(n+2, k)*2^k. - Paul Barry, Jan 23 2006
a(n) = Jacobi_P(n, 2, 0, 3). - Paul Barry, Jan 23 2006
a(n) = (-1)^n*((2*n-1)*LegendreP(n,-3)-LegendreP(n-1,-3))/(2*n+2). - Mark van Hoeij, Oct 31 2011
Recurrence: (n+1)*(6*n-7)*a(n) = (36*n^2-23*n+7)*a(n-1) - (6*n^2-n-21)*a(n-2) + (n-3)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(8+6*sqrt(2))*(3+2*sqrt(2))^n/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 2012
a(n) = hypergeom([-n-1, -n+1], [1], 2). - Peter Luschny, Nov 19 2014
From Peter Bala, Mar 02 2017: (Start)
a(n+1) = 1/2^(n+1) * Sum_{k >= 2} 1/2^k * binomial(n + k, n)*binomial(n + k, n + 2).
(n+1)*(n-1)^2*a(n) = (2*n-1)*(3*n^2 -3*n +1)*a(n-1) - (n-2)*n^2*a(n-2) with a(1) = 1 and a(2) = 7. (End)
a(n) = binomial(2*n, n+1)*hypergeom([-n+1, -n-1], [-2*n], -1). - Detlef Meya, Dec 26 2023
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